How do you find all the zeros of f(x)=6x3+25x2+3x−4?

1 Answer
George C.
Aug 5, 2016

f(x) has zeros: 13, −12 and −4

Explanation:

f(x)=6x3+25x2+3x−4

By the rational root theorem, any rational zeros of f(x) are expressible in the form pq for integers p,q with p a divisor of the constant term −4 and q a divisor of the coefficient 6 of the leading term.

So the only possible rational zeros of f(x) are:

±16,±13,±12,±23,±1,±43,±2,±4

Since the coefficients of f(x) have one change of sign, we can use Descartes' rule of signs to deduce that it has exactly one positive Real zero.

Note that:

f(0)=−4 and f(1)=6+25+3−4=30

So there is a Real root somewhat closer to 0 than to 1.

f(16)=6(1216)+25(136)+3(16)−4

=1+25+18−14436=−10036=−259

f(13)=6(127)+25(19)+3(13)−4

=2+25+9−369=0

So x=13 is a zero and (3x−1) a factor:

6x3+25x2+3x−4=(3x−1)(2x2+9x+4)

We can factor the remaining quadratic using an AC method:

FInd a pair of factors of AC=2⋅4=8 with sum B=9. The pair 8,1 works. Use this pair to split the middle term and factor by grouping:

2x2+9x+4

=2x2+8x+x+4

=2x(x+4)+1(x+4)

=(2x+1)(x+4)

Hence the other two zeros are: x=−12 and x=−4

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